Are De Broglie relations only applicable to particles that have zero potential energy? De Broglie relations are always written as:
$$E=h\nu$$
$$p=\frac{h}{\lambda}$$
However, it doesn't make sense when you have waves that are eigenstates of a Hamiltonian operator with a non-zero potential. That is because that would give us a direct relation between momentum and energy: $\frac{E}{p}=v_p$, where $v_p$ is the phase velocity. This doesn't seem to make sense to me. So, would these equations be the same with the electron of the hydrogen atom, for example?
 A: De Broglie's proposition is a limited-validity rule assigning a travelling wave to a freely moving particle. If we have eigenstates of a Hamiltonian with some potential term, we refer to Schroedinger's theory instead, which describes more complicated situations than de Broglie relations do.
The relation
$$
\nu = \frac{E}{h}
$$
is still valid in Schroedinger's description of Hamiltonian eigenstates, here $\nu$ is frequency of oscillation of function $\Psi$. This reminds of de Broglie's wave, but $\Psi$ of an Hamiltonian eigenstate is usually not a plane wave. It is usually some function concentrated in those regions of configuration space that are probable configurations of the system. For single electron atom, $\Psi$ is concentrated near the nucleus and it is not a travelling wave.
Hence, the relation
$$
\lambda = \frac{h}{p}
$$
is usually no longer applicable in this context, as there is no single momentum $p$ of the electron. Energy eigenstates have definite energy, but usually do not have definite momentum.
